PostgreSQL使用不同的ORDER BY打开

我想运行此查询：

SELECT DISTINCT ON (address_id) purchases.address_id, purchases.*
FROM purchases
WHERE purchases.product_id = 1
ORDER BY purchases.purchased_at DESC

但是我得到这个错误：

PG ::错误：错误：SELECT DISTINCT ON表达式必须匹配初始ORDER BY表达式

添加address_id为第一个ORDER BY表达式可以使错误消失，但是我真的不想添加sorting over address_id。是否可以不按顺序进行address_id？

文档说：

DISTINCT ON（expression [，...]）仅保留给定表达式求值相等的每组行的第一行。[...]注意，除非使用ORDER BY来确保所需的行首先出现，否则每个集合的“第一行”都是不可预测的。[...] DISTINCT ON表达式必须与最左边的ORDER BY表达式匹配。

官方文件

因此，您必须添加 address_id至。

另外，如果您要查找包含每个产品的最新购买商品的完整行，address_id并且结果按purchased_at那么您将尝试解决最大的每组N个问题，可以通过以下方法解决：

适用于大多数DBMS的常规解决方案：

SELECT t1.* FROM purchases t1
JOIN (
    SELECT address_id, max(purchased_at) max_purchased_at
    FROM purchases
    WHERE product_id = 1
    GROUP BY address_id
) t2
ON t1.address_id = t2.address_id AND t1.purchased_at = t2.max_purchased_at
ORDER BY t1.purchased_at DESC

基于@hkf的答案的更面向PostgreSQL的解决方案：

SELECT * FROM (
  SELECT DISTINCT ON (address_id) *
  FROM purchases 
  WHERE product_id = 1
  ORDER BY address_id, purchased_at DESC
) t
ORDER BY purchased_at DESC

在此阐明，扩展和解决的问题：选择按某列排序且在另一列上不同的行

您可以在子查询中按address_id排序，然后在外部查询中按所需内容排序。

SELECT * FROM 
    (SELECT DISTINCT ON (address_id) purchases.address_id, purchases.* 
    FROM "purchases" 
    WHERE "purchases"."product_id" = 1 ORDER BY address_id DESC ) 
ORDER BY purchased_at DESC

一个子查询可以解决这个问题：

SELECT *
FROM  (
    SELECT DISTINCT ON (address_id) *
    FROM   purchases
    WHERE  product_id = 1
    ) p
ORDER  BY purchased_at DESC;

中的前导表达式ORDER BY必须与中的列一致DISTINCT ON，因此您不能按同一列中的不同列进行排序SELECT。

ORDER BY如果要从每个集合中选择特定的行，请仅在子查询中使用附加项：

SELECT *
FROM  (
    SELECT DISTINCT ON (address_id) *
    FROM   purchases
    WHERE  product_id = 1
    ORDER  BY address_id, purchased_at DESC  -- get "latest" row per address_id
    ) p
ORDER  BY purchased_at DESC;

如果purchased_at可以的话NULL，考虑一下DESC NULLS LAST。但是，如果要使用它，请确保匹配您的索引。看到：

• PostgreSQL按datetime asc排序，是否为null？
• 为什么ORDER BY NULLS LAST影响主键上的查询计划？

相关，更多说明：

• 在每个GROUP BY组中选择第一行？
• PostgreSQL按datetime asc排序，是否为null？

窗口函数可以一口气解决这个问题：

SELECT DISTINCT ON (address_id) 
   LAST_VALUE(purchases.address_id) OVER wnd AS address_id
FROM "purchases"
WHERE "purchases"."product_id" = 1
WINDOW wnd AS (
   PARTITION BY address_id ORDER BY purchases.purchased_at DESC
   ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING)

对于使用Flask-SQLAlchemy的任何人来说，这对我都有效

from app import db
from app.models import Purchases
from sqlalchemy.orm import aliased
from sqlalchemy import desc

stmt = Purchases.query.distinct(Purchases.address_id).subquery('purchases')
alias = aliased(Purchases, stmt)
distinct = db.session.query(alias)
distinct.order_by(desc(alias.purchased_at))

您也可以使用group by子句来完成此操作

   SELECT purchases.address_id, purchases.* FROM "purchases"
    WHERE "purchases"."product_id" = 1 GROUP BY address_id,
purchases.purchased_at ORDER purchases.purchased_at DESC